Question 370570
a9 = r + 4 = a1+ (9-1)2 = a1 + 16.  Hence a1 = r - 12.
Thus a3 = r - 12 + 2*2 = r  -12+4 = r - 8, and 
a13 = r - 12 + 12*2 = r - 12 + 24 = r + 12.  The sum of the 3rd and 13th terms is then r - 8 + r + 12 = 2r + 4.